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SCORPION-xisa [38]
3 years ago
9

Solve for x -4(-5x+2)-3x-5=-30

Mathematics
2 answers:
valentina_108 [34]3 years ago
5 0

Answer:

x=-1

Step-by-step explanation:

-4(-5x+2)-3x-5=-30

open the bracket

+20x-8-3x-5=-30

collect like terms

20x-3x-8-5=-30

17x=-30+13

17x= - 17

divide both sides by 17

x= -1

zvonat [6]3 years ago
5 0
Respuesta x=-1

-4(-5x+2)-3x-5=-30

20x-8-3x-5=-30

17x-13=-30
+13=+13
17x=-17

17/17=-17/17
X=-1

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y= -3x/5

Step-by-step explanation:

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An image of a rectangular pyramid is shown below: A right rectangular pyramid is shown. Part A: A cross section of the rectangul
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Answer:

Part a) Rectangle

Part b) Triangle

Step-by-step explanation:

<u><em>The picture of the question in the attached figure N 1</em></u>

Part A) A cross section of the rectangular pyramid is cut with a plane parallel to the base. What is the name of the shape created by the cross section?

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When a geometric plane slices any right pyramid so that the cut is parallel to the plane of the base, the cross section will have the same shape (but not the same size) as the base, So, in the case of a right rectangular pyramid, the cross section is a rectangle

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we know that

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see  the attached figure N 2 to better understand the problem

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3 years ago
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If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

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2 years ago
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